Posts Tagged ‘statistical mechanics’

I’d intended to rework the exam problems over the summer and make that the last update to my stat mech notes. However, I ended up studying world events and some other non-mainstream ideas intensively over the summer, and never got around to that final update.

Since I’m starting a new course (condensed matter) soon, I’ll end up having to focus on that, and have now posted a final version of my notes as is.

Here’s part of a problem from our final exam. I’d intended to redo the whole exam over the summer, but focused my summer study on world events instead. Perhaps I’ll end up eventually doing this, but for now I’ll just post this first part.

Question: Large volume Fermi gas density (2013 final exam pr 1)

Write down the expression for the grand canonical partition function of an ideal three-dimensional Fermi gas with atoms having mass at a temperature and a chemical potential (or equivalently a fugacity ). Consider the high temperature “classical limit” of this ideal gas, where and one gets an effective Boltzmann distribution, and obtain the equation for the density of the particles

by converting momentum sums into integrals. Invert this relationship to find the chemical potential as a function of the density .

Hint: In the limit of a large volume :

Answer

Since it was specified incorrectly in the original problem, let’s start off by verifing the expression for the number of particles (and hence the number density)

Moving on to the problem, we’ve seen that the Fermion grand canonical partition function can be written

In section 6.2 [1] we have the following notation for the sums in the grand partition function . Note that I’ve switched notations from as used in class to as used on our final exam. The text uses a script Q like but with the loop much more disconnected and hard to interpret.

This was shorthand notation for the canonical ensemble, subject to constraints on and

I found this notation pretty confusing, since the normal conventions about what is a dummy index in the various summations do not hold.

The claim of the text (and in class) is that we could write out the grand canonical partition function as

Let’s verify this for a Fermi-Dirac distribution by dispensing with the notational tricks and writing out the original specification of the grand canonical partition function in long form, and compare that to the first few terms of the expansion of eq. 1.0.5.

Let’s consider a specific value of , namely all those values of that apply to . Note that we have only for a Fermi-Dirac sysstem, so this means we can have values of like

Our grand canonical partition function, when written out explicitly, will have the form

Okay, that’s simple enough and really what the primed notation is getting at. Now let’s verify that after simplification this matches up with eq. 1.0.5. Expanding this out a bit we have

This completes the verification of the result as expected. It is definitely a brute force way of doing so, but easy to understand and I found for myself that it removed some of the notation that obfuscated what is really a simple statement.

Once we are comfortable with this Fermi-Dirac expression of the grand canonical partition function, we can then write it in the product form that leads to the sum that we want after taking logs

Consider a Bose gas with particles having no spin and obeying an ultra relativisitic dispersion . Unlike photons or phonons, these particles are {\bf conserved}, and hence we must determine the chemical potential which fixes their density. Working in three dimensions, show whether or not these particles will exhibit Bose condensation, and find if it is nonzero.

Answer

For the number of particles in the gas, as with photons, we still have

As in the discussion of low velocity particles in [1] section 7.1, the ground state term has been split out, before making any continuum approximation of the sum over the energetic states.

Writing

where the number of particles in the ground state is chemical potential and temperature dependent

We proceed with the continuum approximation for the number of particles in the energetic states

So we have

Note that , a fixed number. The key feature of Bose condensation remains. There is a finite limit to the number of particles that can be in the energetic state at a given temperature and volume. Any remaining particles are forced into the ground state.

In general the number of particles in the ground state is

and we will necessarily have particles in this state if

That temperature threshold is the Bose condensation temperature

With , , we have for the ground state average number density

This is plotted in fig. 1.1.

Fig 1.1: Ratio of ground state number density to total number density

From the figure it appears that the notion of any sort of absolute condensation temperature is an approximation. We can start having particles go into the ground state at higher temperatures than , but once the chemical potential starts approaching zero, that temperature for which we start having particles in the ground state approaches . The key takeout idea appears to be, once the temperature does drop below , we necessarily start having a non-zero ground state population, and as the temperature drops more and more, the ratio of the number of particles in the ground state relative to the total approaches unity (all particles are forced into the ground state).

The diffusion constant relation to the probability current is referred to as Fick’s law

with which we can cast the probability diffusion identity into a continuity equation form

In 3D (with the Maxwell distribution frictional term), this takes the form

Maxwell distribution

Add a frictional term to the velocity space diffusion current

For steady state the continity equation leads to

We also find

and identify

Hamilton’s equations

SHO

Quantum energy eigenvalues

Liouville’s theorem

Regardless of whether we have a steady state system, if we sit on a region of phase space volume, the probability density in that neighbourhood will be constant.

Ergodic

A system for which all accessible phase space is swept out by the trajectories. This and Liouville’s threorm allows us to assume that we can treat any given small phase space volume as if it is equally probable to the same time evolved phase space region, and switch to ensemble averaging instead of time averaging.

Question: Low temperature Fermi gas chemical potential

In class, we assumed that was quadratic in as a mechanism to invert this non-linear equation. Without making this quadratic assumption find the lowest order, non-constant approximation for .

Answer

To determine an approximate inversion, let’s start by multiplying eq. 1.0.2 by to non-dimensionalize things

or

If we are looking for an approximation in the neighborhood of , then the LHS factor is approximately one, whereas the fractional difference term is large (with a corresponding requirement for to be small. We must then have

Question: Relativisitic Fermi gas ([1], pr 9.3)

Consider a relativisitic gas of particles of spin obeying Fermi statistics, enclosed in volume , at absolute zero. The energy-momentum relation is

where , and is the rest mass.

Find the Fermi energy at density .

With the pressure defined as the average force per unit area exerted on a perfectly-reflecting wall of the container.

Set up expressions for this in the form of an integral.

Define the internal energy as the average .
Set up expressions for this in the form of an integral.

Show that at low densities, and at high densities. State the criteria for low and high densities.

There may exist a gas of neutrinos (and/or antineutrinos) in the cosmos. (Neutrinos are massless Fermions of spin .) Calculate the Fermi energy (in eV) of such a gas, assuming a density of one particle per .

Attempt exact evaluation of the various integrals.

Answer

We’ve found [3] that the density of states associated with a 3D relativisitic system is

For a given density , we can find the Fermi energy in the same way as we did for the non-relativisitic energies, with the exception that we have to integrate from a lowest energy of instead of (the energy at ). That is

Solving for we have

We’ll see the constant factor above a number of times below and designate it

so that the Fermi energy is

For the pressure calculation, let’s suppose that we have a configuration with a plane in the orientation as in fig. 1.1.

Fig 1.1: Pressure against x,y oriented plane

It’s argued in [4] section 6.4 that the pressure for such a configuration is

where is the number density and is a normalized distribution function for the velocities. The velocity and momentum components are related by the Hamiltonian equations. From the Hamiltonian eq. 1.1 we find \footnote{ Observe that by squaring and summing one can show that this is equivalent to the standard relativisitic momentum .} (for the x-component which is representative)

For we can summarize these velocity-momentum relationships as

Should we attempt to calculate the pressure with this parameterization of the velocity space we end up with convergence problems, and can’t express the results in terms of . Let’s try instead with a distribution over momentum space

Here the momenta have been scaled to have units of energy since we want to express this integral in terms of energy in the end. Our normalized distribution function is

but before evaluating anything, we first want to change our integration variable from momentum to energy. In spherical coordinates our volume element takes the form

Implicit derivatives of

gives us

Our momentum volume element becomes

For our distribution function, we can now write

where is determined by the requirement

The z component of our momentum can be written in spherical coordinates as

Noting that

all the bits come together as

Letting , this is

We could conceivable expand the numerators of each of these integrals in power series, which could then be evaluated as a sum of terms.

Note that above the Fermi energy also has an integral representation

or

Observe that we can use this result to remove the dependence of pressure on this constant

Now for the average energy difference from the rest energy

So the average energy density difference from the rest energy, relative to the rest energy, is

From eq. 1.0.24 and eq. 1.0.26 we have

or

This ratio of integrals is supposed to resolve to 1 and 2 in the low and high density limits. To consider this let’s perform one final non-dimensionalization, writing

The density, pressure, and energy take the form

We can rewrite the square roots in the number density and energy density expressions by expanding out the completion of the square

Expanding the distribution about , we have

allowing us to write, in the low density limit with respect to

Low density result

An exact integration of the various integrals above is possible in terms of special functions. However, that attempt (included below) introduced an erroneous extra factor of . Given that this end result was obtained by tossing all but the lowest order terms in and , let’s try that right from the get go.

For the pressure we have an integrand containing a factor

Our pressure, to lowest order in and is then

Our energy density to lowest order in and from eq. 1.0.33c is

Comparing these, we have

or in this low density limit

High density limit

For the high density limit write , so that the distribution takes the form

This can be approximated by a step function, so that

With a change of variables , we have

Comparing both we have

or

Wow. That’s pretty low!

Pressure integral

Of these the pressure integral is yields directly to Mathematica

where is a modified Bessel function [5] of the second kind as plotted in fig. 1.2.

Fig 1.2: Modified Bessel function of the second kind

Plugging this into the series for the pressure, we have

Plotting the summands for in fig. 1.4 shows that this mix of exponential Bessel and quadratic terms decreases with .

Plotting this sum in fig. 1.3 numerically to 10 terms, shows that we have a function that appears roughly polynomial in and .

Fig 1.3: Pressure to ten terms in z and theta

Fig 1.4: Pressure summands

For small it can be seen graphically that there is very little contribution from anything but the term of this sum. An expansion in series for a few terms in and gives us

This allows a and approximation of the pressure

Number density integral

For the number density, it appears that we can evaluate the integral using integration from parts applied to eq. 1.0.30.30

Expanding in series, gives us

Here the binomial coefficient has the meaning given in the definitions of \statmechchapcite{nonIntegralBinomialSeries}, where for negative integral values of we have

Expanding in series to a couple of orders in and we have

To first order in and this is

which allows a relation to pressure

It’s kind of odd seeming that this is quadratic in temperature. Is there an error?

Energy integral

Starting from eq. 1.0.30c and integrating by parts we have

The integral with the factor of doesn’t have a nice closed form as before (if you consider the a nice closed form), but instead evaluates to a confluent hypergeometric function [6]. That integral is

and looks like fig. 1.5. Series expansion shows that this hypergeometricU function has a singularity at the origin

Fig 1.5: Plot of HypergeometricU, and with theta^5 scaling

so our multiplication by brings us to zero as seen in the plot. Evaluating the complete integral yields the unholy mess

to first order in and this is

Comparing pressure and energy we have for low densities (where )

or

It appears that I’ve picked up an extra factor of somewhere, but at least I’ve got the low density expression. Given that I’ve Taylor expanded everything anyways around and this could likely have been done right from the get go, instead of dragging along the messy geometric integrals. Reworking this part of this problem like that was done above.

One can measure the specific heat in this Bose condensation phenomina for materials such as Helium-4 (spin 0). However, it turns out that Helium-4 is actually quite far from an ideal Bose gas.

Photon gas

A system that is much closer to an ideal Bose gas is that of a gas of photons. To a large extent, photons do not interact with each other. This allows us to calculate black body phenomina and the low temperature (cosmic) background radiation in the universe.

An important distinction between a photon sea and some of these other systems is that the photon number is actually not fixed.

Photon numbers are not “conserved”.

If a photon interacts with an atom, it can impart energy and disappear. An excited atom can emit a photon and change its energy level. In a thermodynamic system we can generally expect that introducing heat will generate more photons, whereas a cold sink will tend to generate fewer photons.

We have a few special details that distinguish photons that we’ll have to consider.

spin 1.

massless, moving at the speed of light.

have two polarization states.

Because we do not have a constraint on the number of particles, we essentially have no chemical potential, even in the grand canonical scheme.

Writing

Our number density, since we have no chemical potential, is of the form

Observe that the average number of photons in this system is temperature dependent. Because this chemical potential is not there, it can be quite easy to work out a number of the thermodynamic results.

Photon average energy density

We’ll now calculate the average energy density of the photons. The energy of a single photon is

so that the average energy density is

Mathematica tells us that this integral is

for an end result of

Phonons and other systems

There is a very similar phenomina in matter. We can discuss lattice vibrations in a solid. These are called phonon modes, and will have the same distribution function where the only difference is that the speed of light is replaced by the speed of the sound wave in the solid. Once we understand the photon system, we are able to look at other Bose distributions such as these phonon systems. We’ll touch on this very briefly next time.